Timeline for About reflexivity of ultrapower
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Mar 5, 2018 at 13:08 | history | edited | Matthew Daws | CC BY-SA 3.0 |
typo
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| Mar 5, 2018 at 12:07 | comment | added | Tomasz Kania | @MatthewDaws, you mean `$E$ is separable' not merely reflexive (this is a typo as you employ separability, of course). | |
| Mar 5, 2018 at 9:54 | comment | added | Matthew Daws | In the case of countably complete ultrafilter, then yes. The same argument I gave for the case when $\cup A_n = I$. | |
| Mar 4, 2018 at 19:59 | comment | added | MSMalekan | @MatthewDaws It is true that if $(A_n)$ is a sequence of subsets of $I$ such that $\bigcup A_n\in\mathcal U$ then at least one of $A_n$'s is in $\mathcal U$ | |
| Mar 4, 2018 at 19:48 | comment | added | MSMalekan | @TomekKania In the mentioned situation, we have also $(\ell^2(E))_\mathcal U\cong\ell^2(E)$, so the reflexivity of spaces are equivalent. | |
| Mar 4, 2018 at 16:50 | history | edited | Matthew Daws | CC BY-SA 3.0 |
Add comment about countably complete case
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| Mar 4, 2018 at 11:22 | comment | added | Tomasz Kania | @MeisamSoleimaniMalekan, if $E$ is separable and $\mathcal U$ happens to be countably complete, then $E\cong E^{\mathcal U}$, so you cannot counclude anything. | |
| Mar 4, 2018 at 9:03 | comment | added | Matthew Daws | That I am not sure about. If $\mc U$ is not countably incomplete, then you are sort of into the realm of set theory, and not analysis any more... | |
| Mar 4, 2018 at 3:48 | comment | added | MSMalekan | How do we answer the above question when $\mathcal U$ is not countably incomplete? | |
| Mar 4, 2018 at 3:45 | vote | accept | MSMalekan | ||
| Mar 3, 2018 at 17:42 | history | answered | Matthew Daws | CC BY-SA 3.0 |